a train starting from rest attains a velocity of 72 km/h in 5 min. assuminf that the acceleration is uniform, find acceleration and distance travelled by it for attaining this velocity
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Firstly,this is a physics question,but fortunately I happen to be good I it.
Final velocity =v=72km/h=20m/s
Initial velocity =u=0
5min=300s
By the first law of motion -
V=u+at
20=0+a*300
20/300=a
1/15=a
By the third law of motion-
2as=ut+1/2at^2
2*1/15*s=0*300+1/2*300*300
2/15s=45000
S=45000*2/15
S=6000m=6km
Final velocity =v=72km/h=20m/s
Initial velocity =u=0
5min=300s
By the first law of motion -
V=u+at
20=0+a*300
20/300=a
1/15=a
By the third law of motion-
2as=ut+1/2at^2
2*1/15*s=0*300+1/2*300*300
2/15s=45000
S=45000*2/15
S=6000m=6km
Similar questions
S=ut +1/2at^2=1/15*1/2*300*300=3000m=3km